Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-q + 10}{-q + 1} \div \dfrac{q - 10}{q^2 + 7q - 8} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-q + 10}{-q + 1} \times \dfrac{q^2 + 7q - 8}{q - 10} $ First factor the quadratic. $p = \dfrac{-q + 10}{-q + 1} \times \dfrac{(q - 1)(q + 8)}{q - 10} $ Then factor out any other terms. $p = \dfrac{-(q - 10)}{-(q - 1)} \times \dfrac{(q - 1)(q + 8)}{q - 10} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -(q - 10) \times (q - 1)(q + 8) } { -(q - 1) \times (q - 10) } $ $p = \dfrac{ -(q - 10)(q - 1)(q + 8)}{ -(q - 1)(q - 10)} $ Notice that $(q - 10)$ and $(q - 1)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -\cancel{(q - 10)}(q - 1)(q + 8)}{ -\cancel{(q - 1)}(q - 10)} $ We are dividing by $q - 1$ , so $q - 1 \neq 0$ Therefore, $q \neq 1$ $p = \dfrac{ -\cancel{(q - 10)}\cancel{(q - 1)}(q + 8)}{ -\cancel{(q - 1)}\cancel{(q - 10)}} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $p = \dfrac{-(q + 8)}{-1} $ $p = q + 8 ; \space q \neq 1 ; \space q \neq 10 $